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Unformatted text preview: Therefore v^(13) = v^(-1) = v' (complex conjugate of v). We know that exp(j*theta)+exp(-j*theta) = 2*cos(theta) exp(j*theta)-exp(-j*theta) = 2*j*sin(theta) and thus v + v^13 = 2*cos(6*pi/7) v - v^13 = 2*j*sin(6*pi/7) (vi) The polynomial is z^5-1 . Based on our discussion, z^5 = 1 has z = 1 as its root, along with four more roots: exp(j*2*k*pi/5) for k=1,2,. ..,4. This is the same as the set w^5=1, w, w^2, w^3, and w^4 Sheet1 Page 2 (vii) The geometric sum formula gives 1 + z + z^2 + . .. + z^(n-1) = (1 - z^n)/(1 - z) for all z except z=1. Setting z=v, we have 1 + w + w^2 + w^3 + w^4 = (1 - w^5)/(1 - w) But w^5 = 1, and thus 1 + w + w^2 + w^3 + w^4 = 0...
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- Spring '08